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neural networks algorithms applications and programming techniques phần 10

là số không. Cơ chế đào tạo các trọng lượng sẽ được mô tả sau. Giả định cuối cùng là giá trị ban đầu cho F là số không. Hãy xem xét những gì xảy ra khi Qn được áp dụng đầu tiên. Cú pháp của câu lệnh ném là: ném ngoại lệ đối tượng, đối tượng ngoại lệ phải là một đối tượng thuộc một trong các lớp con của Throwable. | 358 Spatiotemporal Pattern Classification Figure 9.14 A simple two-node SCAF is shown. is zero. The mechanism for training these weights will be described later. The final assumption is that the initial value for T is zero. Consider what happens when Q11 is applied first. The net input to unit 1 is fl.net u Qll 12 2 - T t 1 o - r i where we have explicitly shown gamma as a function of time. According to Eqs. 9.2 through 9.4 -----axỵ b l - r so x begins to increase since r and . are initially zero. The net input to unit 2 is Ỉ2 net 2 Qll W2iaq T t 0 Xi - T t Thus 2 also begins to rise due to the positive contribution from x Since both x and a 2 are increasing from the start the total activity in the network x X2 increases quickly. Under these conditions T t will begin to increase according to Eq. 9.9 . 9.3 The Sequential Competitive Avalanche Field 359 After a short time we remove Q11 and present Q12- xl will begin to decay but slowly with respect to its rise time. Now we calculate Il net and Z2.net again I .net zl Q 2 WI2.T2 T t 0 0 - T t Ỉ2.net z2 Ql2 m21X r 0 1 X - r t Using Eqs. 9.2 through 9.4 again at cax 1 and 2 6 1 xl D so xl continues to decay but In will continue to rise until 1 xl r t . Figure 9.15 a shows how xl and .7 2 evolve as a function of time. A similar analysis can be used to evaluate the network output for the opposite sequence of input vectors. When Q 2 is presented first X 2 will increase. xl remains at zero since Zi.net -T t and thus .7 1 cax . The total activity in the system is not sufficient to cause T i to rise. When Qn is presented the input to unit 1 is 11 1. Even though Z2 is nonzero the connection weight is zero so Xi does not contribute to the input to unit I. xl begins to rise and r t begins to rise in response to the increasing total activity. In this case r does not increase as much as it did in the first example. Figure 9.15 b shows the behavior of xl andat2 for this example. The values of T t for both cases are shown in Figure .